Question: Divide the following complex numbers. $ \dfrac{-8-8i}{-2-2i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2+2i}$ $ \dfrac{-8-8i}{-2-2i} = \dfrac{-8-8i}{-2-2i} \cdot \dfrac{{-2+2i}}{{-2+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8-8i) \cdot (-2+2i)} {(-2-2i) \cdot (-2+2i)} = \dfrac{(-8-8i) \cdot (-2+2i)} {(-2)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8-8i) \cdot (-2+2i)} {(-2)^2 - (-2i)^2} = $ $ \dfrac{(-8-8i) \cdot (-2+2i)} {4 + 4} = $ $ \dfrac{(-8-8i) \cdot (-2+2i)} {8} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-8i}) \cdot ({-2+2i})} {8} = $ $ \dfrac{{-8} \cdot {(-2)} + {-8} \cdot {(-2) i} + {-8} \cdot {2 i} + {-8} \cdot {2 i^2}} {8} $ Evaluate each product of two numbers. $ \dfrac{16 + 16i - 16i - 16 i^2} {8} $ Finally, simplify the fraction. $ \dfrac{16 + 16i - 16i + 16} {8} = \dfrac{32 + 0i} {8} = 4 $